n moles of a perfect gas undergo a cyclic pro | vedclass

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(D) The total work done in a cyclic process is the sum of work done in each individual process.$1$. For process $A \rightarrow B$ (Isothermal expansio

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$nRT \left(\ln 2-\frac{1}{2}\right)$

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(D) The total work done in a cyclic process is the sum of work done in each individual process.$1$. For process $A \rightarrow B$ (Isothermal expansion):$W_{AB} = nRT \ln \left(\frac{V_{2}}{V_{1}}\right) = nRT \ln \left(\frac{2V_{1}}{V_{1}}\right) = nRT \ln 2$.$2$. For process $B \rightarrow C$ (Isobaric compression):Since the process is isothermal from $A$ to $B$,$P_{1}V_{1} = P_{2}V_{2}$. Given $V_{2} = 2V_{1}$,we have $P_{1}V_{1} = P_{2}(2V_{1})$,which means $P_{2} = \frac{P_{1}}{2}$.$W_{BC} = P_{2}(V_{1} - V_{2}) = P_{2}(V_{1} - 2V_{1}) = -P_{2}V_{1}$.Using the ideal gas law at point $B$,$P_{2}V_{2} = nRT$,so $P_{2}(2V_{1}) = nRT$,which gives $P_{2}V_{1} = \frac{nRT}{2}$.Thus,$W_{BC} = -\frac{nRT}{2}$.$3$. For process $C \rightarrow A$ (Isochoric change):Since the volume is constant $(V_{1})$,the work done $W_{CA} = 0$.Total work done $W_{net} = W_{AB} + W_{BC} + W_{CA} = nRT \ln 2 - \frac{nRT}{2} + 0 = nRT \left(\ln 2 - \frac{1}{2}\right)$.

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